Let $z_1,z_2,z_3,z_4\in\mathbb C$ be distinct. State conditions in terms of computation of complex numbers, which make $z_1,z_2,z_3,z_4$ vertices of a square (in the counterclockwise direction).

Since $z_1,z_2,z_3,z_4$ are going to be vertices of a square, I know I need the distance between $z_1$ and $z_2$ equal to the distance between $z_2$ and $z_3$, and so on... Would a correct way to state this be: \begin{equation} |z_1-z_2|=|z_2-z_3|=|z_3-z_4|=|z_4-z_1| \end{equation}

Another condition I can think of is that the distance between $z_1$ and $z_3$ be equal to the distance of $z_2$ and $z_4$, namely \begin{equation} |z_1-z_3|=|z_2-z_4|=\sqrt{2|z_1-z_2|^2} \end{equation}

Those are the only two conditions I can think of. Are they written correctly, reasonable, and am I missing anything?


  • $\begingroup$ Try using rotation theorem. Hint: Multiplying a complex number by $i$ rotates it by $\frac{\pi}{2}$ with respect to the origin $\endgroup$
    – AvZ
    Jan 21 '15 at 7:57
  • $\begingroup$ Note that if z1= z3 and z2 = z4 you will meet your stated conditions but you will not have a square. I'd add for the second condition that the absolute value must be different from 0 $\endgroup$ Oct 30 '17 at 13:28

So that you don't have to square or take square roots.

Notice that $i(z_2-z_1)=z_3-z_2$ and $i(z_3-z_2)=z_4-z_3$ is enough. Geometrically $a-b$ is a translation of the vector that goes from $b$ to $a$. Geometrically, multiplying by $i$ is a $\pi/2$ rotation counterclockwise.

  • $\begingroup$ Thanks! Should we also impose $z_1-z_4=i(z_4-z_3)$? $\endgroup$ Jan 21 '15 at 15:19
  • $\begingroup$ @EthanAlvaree It is not needed. The conditions already imposed say that three of the sides are equal and perpendicular to each other. The fourth side is forced to be equal too. $\endgroup$
    – Pp..
    Jan 21 '15 at 15:23

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